Question

Problem 2 Electric charge is distributed uniformly along a thin rod of length a, with total charge Q. Find the E field at P. a) If Q is concentrated at the center of the rod. b) If Q uniformly distributed along the road. c)Suppose that the road is infinity long and point P is at the middle of the road at distance b above the road, what will be the E field at point P? 0

Answer 1

Electric charge is distributed uniformly along a thin rod of lenght a, with total charge Q. Find the E field at P.

a) If Q is concentrated at the center of the rod.

Since the load is concentrated in the center of the rod, we can treat this problem as a point load. We will use the following equation:

$\vec{E}=k_{e}\frac{q}{r^{2}}$

where $k_{e}$ is the Coulomb constant, $q$ is the charge and $r$ is the distance between the charge and the point.

$\vec{E}=k_{e}\frac{q}{r^{2}}=k_{e}\frac{Q}{(x+a/2)^{2}}$

b) If Q uniformly distributed along the rod.

We will use the electric field equation due to a continuous load distribution:

$\vec{E}=k_{e}\int \frac{dq}{r^{2}}$

where

$dq=\lambda dr$

where $\lambda$ is the linear charge density.

where the limits of the integral extend from one end of the rod (r = x) to the other end (r = a + x). From the integral you can extract the constants $k_{e}$ and $\lambda$ to obtain

$E=k_{e}\lambda \int_{x}^{a+x}\frac{dr}{r^{2}}$

Solving the integral, we obtain

$E=k_{e}\lambda \left [ -\frac{1}{r} \right ]_{x}^{a+x}=k_{e}\lambda \left ( \frac{1}{x}-\frac{1}{a+x} \right )$

$E=k_{e}\lambda \frac{a}{x(a+x)}$

where $Q=\lambda a$

$E=\frac{k_{e}Q}{x(a+x)}$